Finding Thevenin Equivalent in a Bridge Circuit

[Legion]

I'm not sure how to approach finding the thevenin voltage and resistance with a bridge circuit. I'm trying to solve the circuit below:



The 2k2 is the load. I know the voltage at point B is 9V, but I can't figure out how to find the voltage at point A. Because the bridge is unbalanced I can't use resistor ratios. I also can't use series-parallel rules to collapse the resistances. I tried using voltage loops, ie. the rule that the total voltage in any given circuit loop should be zero, but that just gives me different voltage drops across the resistors depending on the loop I take.

How do I use thevenin with an unbalanced bridge circuit?

[T3sl4co1l]

What do you mean "the voltage at B is 9V"?

With no ground indicated, that voltage is undefined...

Tim

[The Electrician]

There's a trick you can use.  It's explained here:

http://kauko.hallikainen.org/rw/theory/theory6.html

[c4757p]

I also can't use series-parallel rules to collapse the resistances.

No?

Yes you can, you just won't get all the answers at once. Try it, you'll get some information, then you can move on with that.

[Legion]

What do you mean "the voltage at B is 9V"?

With no ground indicated, that voltage is undefined...

Tim

Ground is the negative terminal of the battery. Since point B is common to the anode of the battery, it is 9V above ground.

[IanB]

I also can't use series-parallel rules to collapse the resistances.

No?

Yes you can, you just won't get all the answers at once. Try it, you'll get some information, then you can move on with that.

Surely you can only use series-parallel formulas where two resistors in a circuit can be replaced with a single resistor connecting between the same end points or nodes? In an unbalanced H-bridge I don't see any two resistors that can be replaced with one one while leaving the circuit unchanged?

[Legion]

Surely you can only use series-parallel formulas where two resistors in a circuit can be replaced with a single resistor connecting between the same end points or nodes? In an unbalanced H-bridge I don't see any two resistors that can be replaced with one one while leaving the circuit unchanged?

I can't tell if you're being rhetorical or not. But your statement above is my understanding of why I can't use series-parallel formulas.

[c4757p]

I also can't use series-parallel rules to collapse the resistances.

No?

Yes you can, you just won't get all the answers at once. Try it, you'll get some information, then you can move on with that.

Surely you can only use series-parallel formulas where two resistors in a circuit can be replaced with a single resistor connecting between the same end points or nodes? In an unbalanced H-bridge I don't see any two resistors that can be replaced with one one while leaving the circuit unchanged?

My mistake, I was looking at the one on the right with the missing resistor. 1k+2k7 in series, that in parallel with 3k3, then all in series with 1k5.

[Andy Watson]

Smells like a homework question, but ...
There is no formula (until you've completed the problem) for determining the thevenin equivalent. You will have to make several intermediate steps. The first step I would consider is adding the 1k and 2k7 together.

[Legion]

Smells like a homework question, but ...
There is no formula (until you've completed the problem) for determining the thevenin equivalent. You will have to make several intermediate steps. The first step I would consider is adding the 1k and 2k7 together.

Not a homework question. It's from allaboutcircuits.com.  I'm not in school, I'm trying to teach myself electronics.

[IanB]

My mistake, I was looking at the one on the right with the missing resistor. 1k+2k7 in series, that in parallel with 3k3, then all in series with 1k5.

Unfortunately I couldn't see the original image uploaded by legion before (site blocked). So I was just picturing a regular H-bridge configuration as that is what was described. But now I do see the image after all, and I agree the voltage at point A can easily be found by combining series and parallel resistors.

However, the circuit on the left will have to be brute-forced by writing current or voltage equations and solving them (or by finding the Thevenin equivalent).

[Legion]

I solved it. No real secret to it. I just couldn't visualize it properly. I kept redrawing the circuit until one layout kinda "revealed" the solution.

[Andy Watson]

The easy (easier) way to solve either left or right circuit is to simplify the configuration with star-delta transforms - but that's not really the point of this exercise. Also, star-delta is not usually covered until after thevenin and norton equivalents have been introduced -so I'm guessing that you've not encountered those transforms yet.

[Legion]

I figured I'd throw up the solution in case it helps someone.

To find the thevenin voltage I redrew the circuit like this:


Use series-parallel rules and voltage division to find the voltage at A. Vth = VB - VA.

To find the thevenin resistance I redrew the circuit like this:


Take point A as the anode and B as ground. Use series-parallel rules to get Rth.

Vth = 7.6922V
Rth = 788.64 Ohms

[Excavatoree]

Interesting - two posts in one day where I've realized I've forgotten the delta-wye transformation formulas.

[IanB]

Interesting - two posts in one day where I've realized I've forgotten the delta-wye transformation formulas.

I never particularly remembered them, but I can always work them out if I need to know them. Just recalling that "there is such a transformation" is as much knowledge as you ever really need.

[The Electrician]

I figured I'd throw up the solution in case it helps someone.

To find the thevenin voltage I redrew the circuit like this:


Use series-parallel rules and voltage division to find the voltage at A. Vth = VB - VA.

To find the thevenin resistance I redrew the circuit like this:


Take point A as the anode and B as ground. Use series-parallel rules to get Rth.

Vth = 7.6922V
Rth = 788.64 Ohms

What software did you to make these images, please?

[Legion]

What software did you to make these images, please?

Photoshop. But I wouldn't recommend it. It's really slow. I should probably get some schematic drawing software.